Mathematics — CBSE Class 10 board question

Given: $\sin(2A + B) = \dfrac{\sqrt{3}}{2}$ and $\tan(A + 2B) = 1$

From the standard table:
$$\sin 60° = \frac{\sqrt{3}}{2} \Rightarrow 2A + B = 60° \quad \cdots(1)$$
$$\tan 45° = 1 \Rightarrow A + 2B = 45° \quad \cdots(2)$$

Solving (1) and (2):

From (1): $2A + B = 60°$
From (2): $A + 2B = 45°$

Multiply (2) by 2: $2A + 4B = 90°$

Subtract (1): $3B = 30° \Rightarrow B = 10°$

Substitute in (1): $2A + 10° = 60° \Rightarrow A = 25°$

Now, $4A - 7B = 4(25°) - 7(10°) = 100° - 70° = 30°$

$$\cot(4A - 7B) = \cot 30° = \sqrt{3}$$

Source: Chapter 8, Section 8.3 (Trigonometric Ratios of Specific Angles)

---

• The key step is recognising standard values: $\sin 60° = \frac{\sqrt{3}}{2}$ and $\tan 45° = 1$ from the table, so you can directly equate the compound angles to 60° and 45° respectively.
• Solve the resulting pair of linear equations carefully — a common error is subtraction mistakes.
• Finally, substitute to find $4A - 7B$ and look up $\cot 30° = \sqrt{3}$ from the standard table.
• Examiners award marks for: correct identification of angles (1 mark), solving for A and B (1 mark), and final answer (1 mark).